question_answer

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: \[{{R}_{1}}=15\Omega ,{{R}_{2}}=10\Omega ,{{R}_{3}}=20\Omega ,{{R}_{4}}=5\Omega ,\] \[{{R}_{5}}=25\Omega ,{{R}_{6}}=30\Omega ,E=15V\]             [JEE Main 8-4-2019 Afternoon]

A) 7/18                

B) 13/24

C) 9/32

D)   20/3

Correct Answer: C

Solution :

\[{{R}_{eq}}=15+\frac{25}{3}+30=\frac{45+25+90}{3}=\frac{160}{3}\] \[I=\frac{E}{{{R}_{eq}}}=\frac{15\times 3}{160}=\frac{9}{32}amp.\]